# Understanding Why $\lvert{x}\rvert < \min\{1,\delta\}$ in Spivak's Solutions.

I am working my way through Spivak. I am not incredibly confident in my solution to the problem, and the solutions in the back of the book gives a subtlety that I don't have in my own proof. I'm wondering why.

Claim: $$\lim_{x \to 0} f(x) = \lim_{x \to 0} f(x^3).$$

My Proof: First, suppose $$\lim_{x \to 0} f(x) = L$$. Then, by definition, $$\forall \epsilon, \exists \delta$$ such that $$\forall x, \lvert{x - 0}\rvert < \delta \implies \lvert{f(x) - L}\rvert < \epsilon$$. Since we may choose any $$\delta \in \mathbb{R}^+$$, consider $$\lvert{x}\rvert < \sqrt[3]\delta$$. Then $$\lvert{x^3}\rvert < \delta \implies \lvert{f(x^3) - L}\rvert < \epsilon$$. Thus, $$\lim_{x\to 0} f(x^3) = L$$.

Conversely, suppose $$\lim_{x \to 0} f(x^3) = L$$. Then, again by definition, $$\forall \epsilon, \exists \delta$$ such that $$\forall x, \lvert{x - 0}\rvert < \delta \implies \lvert{f(x^3) - L}\rvert < \epsilon$$. If $$\lvert{x}\rvert < \delta^3$$, then $$\lvert{\sqrt[3]{x}}\rvert < \delta \implies \lvert{f(\sqrt[3]{x}^3}) - L\rvert < \epsilon$$. Thus, $$\lim_{x\to 0} f(x) = L$$.

$$\blacksquare$$

Since this is the chapter on limits, the proof is written very pedantically. However, the crux of the argument for each case is:

1. Since we may choose any $$\delta \in \mathbb{R}^+$$, consider $$\lvert{x}\rvert < \sqrt[3]\delta$$.
2. Since $$\lvert{x}\rvert < \delta$$, we can simply choose $$\delta^3$$ such that $$\lvert{x}\rvert < \delta^3 \implies \lvert{\sqrt[3]{x}}\rvert < \delta$$.

My Question

Spivak's Solutions has the following as justification for the first case:

Then if $$0 < \lvert{x}\rvert < \min{(1, \delta)}$$, we have $$0 < \lvert{x^3}\rvert < \delta$$.

I don't understand why Spivak shows that $$\delta < \min{(1, \delta)}$$, or why knowing this we can deduce $$0 < \lvert{x^3}\rvert < \delta$$.

• If $|x|<1$, then $|x^3|<|x|$. Commented May 3, 2021 at 2:07
• ah got it. thanks @Jonah Commented May 3, 2021 at 2:10

$$|x|<\min (1,\delta) \iff |x|<1 \text{ and } |x|<\delta$$
If $$\delta<1,$$ then $$|x|^3<\delta^3<\delta<1.$$
If $$1<\delta,$$ then $$|x|^3<1<\delta<\delta^3.$$
Otherwise, if $$|x|>1$$, then $$|x|<\delta$$ does not imply $$|x|^3<\delta.$$